Example 1
ΔV = 1.8×10⁻³ m³.
Volumetric Expansion Calculator
Enter γ, the starting volume, and the temperature rise. You get how much the volume grows. At 3.6e-5, 1 m³ and 50 K you get 0.0018 m³. For steel that is roughly 3α.
Linear: Δl = α·l₀·ΔT. Heat at the same ΔT: Q = c·m·ΔT.
Result
γ (1/K), Volume V₀ (m³, SI) and ΔT (K). The result shows up here.
How it works
Volume grows with temperature: ΔV = γ·V₀·ΔT. Final V = V₀ + ΔV. The calculator example: γ = 0.000036, V₀ = 1 m³, ΔT = 50 gives ΔV = 0.0018 m³. That is 1.8 liters per cubic meter, small on that scale, visible on a tank.
For solids γ ≈ 3α, because three dimensions grow. Steel α = 12e-6 gives γ ≈ 36e-6. If you know α from the linear page, you can estimate γ and come back here. Liquids have their own γ from a table, not 3α of steel. Water near 20 °C is on the order of 2e-4 1/K.
Keep V₀ in cubic meters. One liter is 0.001 m³. If you type 1 thinking of a liter, ΔV comes out a thousand times too large. γ is in 1/K. ΔT is again a difference: 20 K, not 293.
At V₀ = 0.001 m³, γ = 3.6e-5 and ΔT = 20 K you get ΔV = 7.2e-7 m³, about 0.72 ml. With the same data and 50 K it is 1.8 ml. A negative ΔT means contraction.
This formula does not give heat. Energy for the same temperature jump is Q = c m ΔT. Here you only get volume geometry.
Type γ, V₀ and ΔT, then Calculate. 0.000036, 1 and 50 should come back as 0.0018 m³. A comma in 0.000036 parses the same as a period.
How to use
- Type γ. For steel about 0.000036, that is 3.6e-5, roughly 3α.
- Type V₀ in m³. One liter is 0.001, not 1.
- Type ΔT as a rise in kelvin, for example 50. Do not add 273.
- Click Calculate. 0.000036, 1 m³ and 50 K give 0.0018 m³.
- For α in one dimension, open linear expansion. For energy, open Q = c m ΔT.
Formula
ΔV = γ · V0 · ΔT
For solids γ ≈ 3α. V₀ in m3.
Letters in ΔV = γ·V₀·ΔT
Volume grows as ΔV = γ·V₀·ΔT. At γ = 3.6×10⁻⁵, 1 m³ and 50 K you get 0.0018 m³, 1.8 liters per cubic meter.
- ΔV
- Volume increase in m³. 0.000036 × 1 × 50 = 0.0018 m³ on the page example.
- γ
- Volume coefficient in 1/K. For steel about 3.6×10⁻⁵, roughly 3α from 12×10⁻⁶.
- V₀
- Starting volume in m³. A one is a cubic meter; type a liter as 0.001, or ΔV is a thousand times too large.
- ΔT
- Rise in kelvin, the same 50 K as on the bar, except here the volume swells, not the length.
Real-life examples
Example 2
α = 1.2×10⁻⁵ → γ ≈ 3.6×10⁻⁵, V₀ = 1, ΔT = 50.
Example 3
V₀ = 0.001, γ = 3.6×10⁻⁵, ΔT = 50.
Example 4
V₀ = 1, ΔT = 40.
Example 5
V₀ = 0.5, ΔT = 80.
Example 6
V₀ = 2, γ = 3.6×10⁻⁵, ΔT = 30.
Example 7
V₀ = 1, ΔT = −25.
Example 8
V₀ = 0.0001, ΔT = 100.
Example 9
γ = 2×10⁻⁴, V₀ = 0.001, ΔT = 20.
Example 10
V₀ = 10, γ = 3.6×10⁻⁵, ΔT = 20.
Ways to use this calculator
- From steel α = 12e-6 you make γ ≈ 36e-6 and compute the rise of a cubic meter.
- You convert a liter to 0.001 m³ so ΔV is not a thousand times too large.
Frequently asked questions
How much ΔV at γ = 3.6e-5, V₀ = 1 m³ and ΔT = 50 K?
ΔV = 0.0018 m³. Final V = 1.0018 m³ after that 50 K rise.
How much at V₀ = 0.001 m³, γ = 3.6e-5 and ΔT = 20 K?
ΔV = 7.2e-7 m³, about 0.72 ml. At 50 K it is 1.8 ml.
Why is γ ≈ 3α at α = 12e-6?
Three dimensions. 3 × 12e-6 = 36e-6 1/K. Liquids have their own γ, not 3α of steel.
Is 1 in the V₀ field a liter?
No. The calculator needs m³. Type a liter as 0.001. A one is already a full cubic meter.
Can ΔT be negative?
Yes. Then ΔV is negative: the volume shrinks on cooling.
What γ does water have near 20 °C?
On the order of 2e-4 1/K, not 3α of steel. Take the constant from a liquid table.
Does this formula compute joules?
No. Heat for that ΔT lives on Q = c m ΔT. Here only ΔV.
Where do I compute a bar growing longer?
On linear expansion, Δl = α l₀ ΔT. From there you take α into 3α.
Does 3.6e-5 work?
Yes. 3.6e-5, 3.6×10⁻⁵ and 0.000036 are the same γ.
What unit is the result?
Cubic meters. 0.0018 m³ is 1.8 liters. The calculator does not convert V₀ to gallons inside the formula.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.